A body of mass (2M) splits into four masses (m, M - m, m, M - m}, which are rearranged to form a square as shown in the figure. The ratio of M/m for which, the gravitational potential energy of the system becomes maximum is x: 1. The value of x is............. Numerical answer type. Work it out, then open the answer.

Answer: 2. Sides (4 pairs of m and M-m, distance d) and diagonals ( m – m and (M-m) – (M-m), distance √(2)d): U = -G/d[4m(M-m) + (m² + (M-m)²)/(√(2))] Extremum: dU/dm = 0 ⇒ 4(M-2m) + (4m - 2M)/(√(2)) = 0 ⇒ (M-2m)(4-√(2)) = 0 M = 2m, at which |U| is greatest (the intended 'maximum'). So M/m = 2, i.e. x = 2.

A body of mass (2M) splits into four masses (m, M - m, m, M - m}… | JEE Main 2021 (27 Aug, Shift 1)

Q19JEE Main 2021 · 27 AUG · SHIFT 1Difficult

Gravitational Potential & Energy

A body of mass (2M) splits into four masses (m, M $-$ m, m, M $-$ m}, which are rearranged to form a square as shown in the figure. The ratio of $\frac{M}{m}$ for which, the gravitational potential energy of the system becomes maximum is x: 1. The value of x is .............

Numerical answer type. Work it out, then open the answer.

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Answer: 2

Sides (4 pairs of $m$ and $M-m$ , distance $d$ ) and diagonals ( $m$ – $m$ and $(M-m)$ – $(M-m)$ , distance $\sqrt{2}d$ ):
$U = -\frac{G}{d}\left[4m(M-m) + \frac{m^2 + (M-m)^2}{\sqrt{2}}\right]$
Extremum: $\frac{dU}{dm} = 0 \Rightarrow 4(M-2m) + \frac{4m - 2M}{\sqrt{2}} = 0 \Rightarrow (M-2m)(4-\sqrt{2}) = 0$
$M = 2m$ , at which $|U|$ is greatest (the intended 'maximum').
So $\frac{M}{m} = 2$ , i.e. $x = 2$ .

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Gravitational Potential & Energy question from JEE Main 2021 (27 Aug, Shift 1)

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